Question

Let $[x]$ denote the greatest integer $\le x$. Consider the function $f(x)=\max \left\{x^2,1+[x]\right\}$. Then the value of the integral $\underset{0}{\overset{2}{\int }}f(x)dx$ is

• A. $\frac{5+4\sqrt{2}}{3}$
• B. $\frac{8+4\sqrt{2}}{3}$
• C. $\frac{1+5\sqrt{2}}{3}$
• D. $\frac{4+5\sqrt{2}}{3}$

Answer 1

Answer: (A)

$A=\underset{0}{\overset{1}{\int }}1\cdot dx+\underset{1}{\overset{\sqrt{2}}{\int }}2dx+\underset{\sqrt{2}}{\overset{2}{\int }}{x}^{2}dx$
$=1+2\sqrt{2}-2+\frac{8}{3}-\frac{2\sqrt{2}}{3}$
$=\frac{5}{3}+\frac{4\sqrt{2}}{3}$